[英]Select Inside Insert Statement - MySQL Cursors
我嘗試了一些,但找不到解決方案,以某種方式設法獲得了這個結果。
這是查詢:
DELIMITER ##
CREATE PROCEDURE test1(start_date DATE,end_date DATE)
BEGIN
DECLARE done INT DEFAULT FALSE;
DECLARE a INT;
DECLARE present INT;
DECLARE total INT;
-- Declare the cursor
DECLARE id CURSOR
FOR
SELECT staff_id FROM ost_staff;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = TRUE;
-- Open the cursor
DROP TEMPORARY TABLE IF EXISTS reports;
CREATE TEMPORARY TABLE IF NOT EXISTS reports
(
staff_id INT(10),
present INT(10),
total INT(10)
);
OPEN id;
read_loop: LOOP
FETCH id INTO a;
IF done THEN
LEAVE read_loop;
END IF;
INSERT INTO reports(staff_id,present,total)
SELECT (COUNT(I.interval_start)) AS present, DATEDIFF(DATE_ADD(end_date,INTERVAL 1 DAY),start_date) AS total
FROM effort_frequency E
RIGHT OUTER JOIN time_intervals I ON I.interval_start = E.log_date
AND E.staffid=a AND E.log_date BETWEEN start_date AND end_date
LEFT OUTER JOIN ost_holidays H ON H.holiday_date = I.interval_start
WHERE DATE_FORMAT(I.interval_start,'%a') = 'Sun' OR H.holiday_date = I.interval_start OR E.total_effortspent IS NOT NULL;
-- Close the cursor
END LOOP;
CLOSE id;
END ##
我得到以下結果:
+----------+-----------------+
| staff_id | present | total |
+----------+---------+-------+
| (NULL) | 23 | 24 |
| (NULL) | 22 | 24 |
+----------+---------+-------+
我的staff_id為(NULL),如何獲得staff_id? 我嘗試在insert語句中使用聲明的變量'a',但是那時我只有staff_id,沒有得到其他2個字段,我無法從insert語句中的select里面得到staff_id,因為存在一些問題。
現在,我需要將變量'a'中的staff_id插入該臨時表中。
注意:我真的是這個存儲過程的新手,但是直到現在仍然可以通過某種方式進行管理,如果我能詳細了解如何使用“插入內部選擇”(包括針對此問題的解決方案),那就很好了。
嘗試這個 -
SELECT a, (COUNT(I.interval_start)) AS present, DATEDIFF(DATE_ADD(end_date,INTERVAL 1 DAY),start_date) AS total
INSERT
需要三個字段,但是SELECT
語句僅選擇兩個: present
和total
。
嘗試:
SELECT E.staffid, (COUNT(I.interval_start))
AS present, DATEDIFF(DATE_ADD(end_date,INTERVAL 1 DAY),start_date) AS total
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